is this question algebra?
- Thread starter moffas01
- Start date
D
Deleted member 4993
Guest
You can solve this problem without algebra.
Simply count it out:
chocolate, strawberry, vanilla
chocolate, strawberry, raspberry
chocolate, vanilla and raspberry
and so on....
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,087
Algebra is not a kind of question, but one of many tools you can use to solve a problem. So a (word) problem can't be classified as an algebra problem in itself.
But this question does fall into the domain of combinatorics, the study of counting. As such, it can be solved by several methods, including mere listing and counting, the fundamental counting principle, and the "nCr" function ("n choose r"). This subject is often taught along with probability, so questions about it often go there.
The hard part often is to define exactly what is being asked. In this case, can you include more than one scoop of the same flavor? Does the order of the scoops matter? That greatly changes the nature of the problem!
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
Suppose you walked into that shop and they handed you a card with four boxes on it. Each box represented a flavor with a picture and a name. To order you are to place three x's on the card, an x to indicate which flavor(s) you want. I would want one vanilla & two raspberry.
The number of ways to place \(N\) identical objects (choices) into \(K\) distinct cells (flavors) is \(\dbinom{N+K-1}{N}=\dfrac{(N+K-1)!}{N!(K-1)!}\)
In this case \(\dfrac{6!}{(3!)(3!)}\).